Class Solution

  • All Implemented Interfaces:

    
    public final class Solution
    
                        

    403 - Frog Jump\.

    Hard

    A frog is crossing a river. The river is divided into some number of units, and at each unit, there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water.

    Given a list of stones' positions (in units) in sorted ascending order , determine if the frog can cross the river by landing on the last stone. Initially, the frog is on the first stone and assumes the first jump must be 1 unit.

    If the frog's last jump was k units, its next jump must be either k - 1, k, or k + 1 units. The frog can only jump in the forward direction.

    Example 1:

    Input: stones = 0,1,3,5,6,8,12,17

    Output: true

    Explanation: The frog can jump to the last stone by jumping 1 unit to the 2nd stone, then 2 units to the 3rd stone, then 2 units to the 4th stone, then 3 units to the 6th stone, 4 units to the 7th stone, and 5 units to the 8th stone.

    Example 2:

    Input: stones = 0,1,2,3,4,8,9,11

    Output: false

    Explanation: There is no way to jump to the last stone as the gap between the 5th and 6th stone is too large.

    Constraints:

    • 2 <= stones.length <= 2000

    • <code>0 <= stonesi<= 2<sup>31</sup> - 1</code>

    • stones[0] == 0

    • stones is sorted in a strictly increasing order.

    • Nested Class Summary

      Nested Classes 
      Modifier and Type Class Description
    • Field Summary

      Fields 
      Modifier and Type Field Description
    • Constructor Summary

      Constructors 
      Constructor Description
      Solution()
    • Enum Constant Summary

      Enum Constants 
      Enum Constant Description
    • Method Summary

      Modifier and Type Method Description
      final Boolean canCross(IntArray stones)
      • Methods inherited from class java.lang.Object

        clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait